Improper integrals over infinite intervals

Warm-up

Three quick questions from earlier lessons.

Discover

$\displaystyle\int_1^b \dfrac{dx}{x^2} = 1 - \dfrac{1}{b}$ . For $b = 10$ , $100$ , $1000$ this is $0.9$ , $0.99$ , $0.999$ . What happens as $b \to \infty$ ?

It approaches 1
It grows without bound

Key idea

An infinite interval is handled with a limit.

\displaystyle\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx

. If the limit is a number, the integral converges to it; otherwise it diverges.

Worked example

$\displaystyle\int_0^{\infty} e^{-2x}\,dx$ .

\displaystyle\int_0^b e^{-2x}\,dx = \dfrac{1 - e^{-2b}}{2}

Integrate to b.

\displaystyle\lim_{b \to \infty} \dfrac{1 - e^{-2b}}{2} = \dfrac{1}{2}

e^{−2b} → 0.

Converges to

\tfrac{1}{2}

A finite limit.

Your turn, guided

Your turn: $\displaystyle\int_1^{\infty} \dfrac{dx}{x^3}$ .

= \tfrac{1}{2}

[−1/(2x²)] from 1 to b → 0 + ½.

Practice ★

Evaluate $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 4 } }$ .
Evaluate $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 3 } }$ .

Evaluate $\displaystyle\int_0^{\infty} e^{-4x}\,dx$ .
Evaluate $\displaystyle\int_0^{\infty} e^{-3x}\,dx$ .

Watch out!

Practice ★★

What is $\displaystyle\int_{ 9 }^{\infty} \dfrac{dx}{x}$ ?
1. It diverges: $\ln b \to \infty$
2. It converges, because $\dfrac{1}{x} \to 0$
3. It equals $\ln 9$

Boss problem. Evaluate $\displaystyle\int_0^{\infty} x e^{-1x}\,dx$ .

Exit ticket

Three questions to check you have got it.

Evaluate $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 4 } }$ .
Evaluate $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 3 } }$ .
Evaluate $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 2 } }$ .

Summary card

Infinite intervals

Replace $\infty$ by $b$ .
Integrate.
Take $\lim_{b \to \infty}$ : a number means convergence.

Watch out: $f \to 0$ does not guarantee convergence.

End of the lesson

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